prove that the product of 4 consecutive natural numbers is not a perfect cube?

Answer 1

Here is one way to show the result. Let your 4 consecutive natural numbers be given by n, n+1, n+2, n+3 and suppose that their product is a cube.

First, note that, for any prime number p>3 dividing the product of these, p can divide only one of the terms. This is true because the numbers under consideration are consecutive. We can rename the 4 given numbers if necessary to say that n and n+2 are odd. 3 divides one of these (it cannot divide both), or it doesn't. It could divide it to the first, second, or third power (modulo 3). If it divides it to the third power (or not at all), then both n and n+2 are cubes (by the comment above about primes in the factorization). If it divides it to the second power, then 3*n*(n+2) is a cube. If it divides it to the first power, then 9*n*(n+2) is a cube. So there are three cases.

Suppose that both n and n+2 are cubes. That is, n*(n+2)=x^3 for some natural number x. Completing the square on the left hand side and letting y=n+1, this is equivalent to y^2=x^3+1. This is an elliptic curve. It turns out that this curve has only finitely many rational points on it, let alone integer points. All such points are:

(2,3) , (0,1) , (-1,0) , (0,-1) , (2,-3).

Recall that we had n an odd number. Translating these points back to the original question gives values of n equal to 2, 0, -1, -2, -4. None of these are positive odd integers.

Now suppose that 3*n*(n+2)=q^3. Again, completing the square and letting z=n+1 gives 3*z^2=q^3+3. Multiplying through by 3^3 and letting y=9*z, x=3*q gives y^2=x^3+81. Again, this is an elliptic curve. It turns out that this curve has only the following rational points on it:

(0,9) , (0,-9).

Translating these back to the original problem gives n=0,-2. Neither of these are odd natural numbers.

Finally, suppose that 9*n*(n+2)=x^3. Completing the square and letting z=n+1 gives 9*z^2=x^3+9. Letting y=3*z gives y^2=x^3+9. This is another elliptic curve. It turns out that there are infinitely many rational points on this curve! But the only integer points are (unless I made a mistake in the calculation):

(0,3) , (0,-3) , (6,15) , (6,-15) , (3,6) , (3,-6) , (-2, 1) , (-2, 1) , (40, 253) , (40, -253).

Translating these back give n=0,-2,4,-6, 1, -3. Only n=1 is an odd natural number, and it gives a possible solution of n=1, n+2=3. So the possible sequence of four natural numbers is only 1*2*3*4, which is not a cube.

These are all cases, so we have a contradiction. Therefore, the product of any 4 consecutive natural numbers can never be a cube.

Note that these calculations could have been done without using elliptic curve methods (possibly). Instead, studying the arithmetic in various quadratic extensions of the rational numbers should lead to the same result.

Answer 2

I don't know the answer off the top of my head, but I know how I'd attack it. A positive integer is a perfect cube if and only if:

When you factor it uniquely into primes, each of the exponents is divisible by 3.

So look at the sequence of four consecutive integers. Two will be odd, one will be even but not divisible by 2^2, and one will be divisible by 2^x for some x>1. So x+1 must be divisible by 3.

Skipping over 3, either one of the integers is divisible by 5 or none is. Well, if there's one divisible by 5, its factorization must include a power of 5 that's divisible by 3 ...

Ya know, this isn't getting anywhere.

The next attack that comes to mind is to write it as, say (x-1)x(x+1)(x+2) = say (x^3 - x)(x + 2), and look for inspiration.

Hmm. I'm stumped at the moment.

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EDIT: In the answer below, there seems to have been an implied "Without loss of generality, the first of the four terms is odd." I guess that's OK, since the proof isn't affected by whether the numbers are positive or negative, but it still seemed a little quick. -- Oh, that was a slight typo. What really was meant, I think, was a focus on the two odd terms. The reason they were chosen to be odd is that, when the powers of 3 are removed, what's left over have to be perfect cubes.


Anyhow, I imagine a more elementary proof was desired ...

Answer 3

by taking examples
1*2*3*4 is not a perfect cube...

Answer 4

0 x 1 x 2 x 3 = 0
0^2 = 0

The statement is false.

Answer 5

if we take 1,2,3,4 as our natural numbers, the answer is 24 and it is not a perfect cube.
5,6,7,8=1680=noetc.

Answer 6

Dear jimmye
0 is not natural.

Answer 7

Its easy man!!! Just take examples!!!!1

Answer 8

No, I'd rather you prove it. We'll wait.